Particle Accelerator

Guru Vishnu

04:04

Graph for capacitor charging and discharging at the same time: desmos.com/calculator/qcworg1e5r

JR0004 - HCV-32-Ex-83

desmos.com/calculator/rlrw079jak

Guru Vishnu

05:54

JR0004 - HCV-32-Ex-83

Question:
A capacitor of capacitance $C$ is given a charge $Q$. At $t=0$, it is connected to an uncharged capacitor of equal capacitance through a resistance $R$. Find the charge on the second capacitor as a function of time.

@JohnRennie: Hi sir. Good morning :-)

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

@JohnRennie Are you free now sir? I need some help in the above question.

John Rennie

You need to write an expression for $dQ/dt$.

Suppose the initial change on the capacitor is $Q_0$, then the voltage is $V_0 = Q_0/C$. Yes?

Guru Vishnu

@JohnRennie But here there are two capacitors involved. One is charging and the other is discharging. In an ordinary circuit with only one capacitor I know the expressions for the charge as a function of time for both charging and discharging. But here I don't understand how to relate the effect one capacitor has on the other, sir.

@JohnRennie Yes sir.

John Rennie

Now suppose a charge $Q$ has flowed off the first capacitor onto the second, so the charge on the first capacitor is $Q_0-Q$ and the charge on the second capacitor is $Q$.

Guru Vishnu

06:02

@JohnRennie Yes sir.

John Rennie

Then the voltage on the first capacitor is $(Q_0-Q)/C$ and the voltage on the second capacitor is $Q/C$.

Can you see where I'm going with this?

Guru Vishnu

@JohnRennie Yes sir. But the presence of resistor on only one branch causes some lack of symmetry. Does it have any role in the final answer?

John Rennie

The potential difference between the two capacitors is $\Delta V = V_1 - V_2 = (Q_0-Q)/C - Q/C = (Q_0 - 2Q)/C$.

And this potential difference is across the resistor, so the current is $dQ/dt = \Delta V/R = (Q_0-2Q)/(RC)$

Guru Vishnu

@JohnRennie: Ok sir.

Is the next step is to apply Kirchhoff's voltage/loop law?

John Rennie

No. You've now got a diffarential equation:

$$ \frac{dQ}{dt} = \frac{Q_0-2Q}{RC} $$

Just solve it for $Q(t)$

Guru Vishnu

06:10

@JohnRennie Ok sir. I'll do it and ask if I have any doubts. Thank you.

John Rennie

@GuruVishnu are you happy you understand what I did and how I got this equation?

You'll see a lot of questions like this so you need to be sure you understand the approach for solving them.

Guru Vishnu

@JohnRennie I'm happy but still the unsymmetrical condition gives some trouble.

The presence of resistor on only one branch would differentially cause drag in one wire more than the other connecting the plates of the two capacitors.

Did I make it clear how the system is unsymmetrical sir? Or may I draw a diagram?

John Rennie

That's the circuit.

Guru Vishnu

Exactly yes sir.

John Rennie

The first capacitor starts with a charge $Q_0$ and charge is going to flow round the circuit until both capacitors end up with a charge $Q_0/2$.

Guru Vishnu

06:15

Yes sir.

John Rennie

I don't see what the problem is ...

Guru Vishnu

@JohnRennie Resistor retards the flow of electrons. Am I right, sir?

John Rennie

I wouldn't put it that way. The current through a resistor is I = V/R. That's all you need to know.

Guru Vishnu

@JohnRennie Ok sir. Is it ok to assume two resistances of R/2 each in two branches satisfying the symmetric condition?

John Rennie

Why would you do that?

Guru Vishnu

06:18

@JohnRennie Just for my satisfaction to see the system as symmetric :-) Also to know whether is it ok to swap the order of capacitors and resistors in a closed loop.

But the second reason is the more important one, sir.

John Rennie

If you do that you make the calculation more complicated because then you have two resistors each with a potential difference across them.

Guru Vishnu

@JohnRennie I agree sir. But will the final result vary on this fact? Because, as per our previous discussions a zero resistance wire causes a lot of trouble logically.

So I thought it would be better to assume some resistance in both wires. That's why I thought of sharing the total resistance with the other branch.

John Rennie

The wire connecting the bottom plates of the capacitors just sets them to be the same potential, so the potential difference we calculated is then between the top plates of the capacitors i.e. across the resistor.

Guru Vishnu

@JohnRennie Yes sir. But in reality, same amount of current must flow through the wires in order to maintain the same amount of charges on the opposite plates of the capacitor and this is why I'm having some trouble understanding the unequal resistance distribution.

John Rennie

Current is like a fluid - like water flowing in a pipe. The current has to be the same everywhere so if you restrict the flow at one point you reduce the flow everywhere.

Guru Vishnu

06:30

Fine sir. Doesn't this call for same resistance on both branches?
Using your analogy:
Speed of flow is constant only if the area of cross section is uniform at constant volume flow rate. But it varies drastically if the area of cross section is different at different regions. And the un-symmetrical condition is similar to that of here, sir.

John Rennie

Not speed of flow, volumetric flow rate. The current is like a volumetric flow rate.

i.e. number of electrons per second passing any point in the circuit.

Guru Vishnu

@JohnRennie Ok sir. Now understood. So irrespective of the resistances of the two branches, the effect on one branch is equally transmitted to the other one. Sounds similar to Pascal's law in fluids :-)

John Rennie

> Sounds similar to Pascal's law in fluids

Yes.

Circuits are remarkably similar to fluids.

Hydraulic analogy

The electronic–hydraulic analogy (derisively referred to as the drain-pipe theory by Oliver Lodge) is the most widely used analogy for "electron fluid" in a metal conductor. Since electric current is invisible and the processes at play in electronics are often difficult to demonstrate, the various electronic components are represented by hydraulic equivalents. Electricity (as well as heat) was originally understood to be a kind of fluid, and the names of certain electric quantities (such as current) are derived from hydraulic equivalents. As with all analogies, it demands an intuitive a...

Guru Vishnu

@JohnRennie Cool :-)

Guru Vishnu

07:09

Thank you sir.

Guru Vishnu

07:22

And solving the differential equation, I obtained the expression:
$$Q(t)=\frac Q 2 (1-e^{-2t/RC})$$

End of #JR0004 - HCV-32-Ex-83

John Rennie

Cool :-)

Guru Vishnu

@JohnRennie Thank you sir. But I missed this message in the bookmark.

Guru Vishnu

08:24

@JohnRennie: Are you free now sir?

Or please reply when you find time: Can we assume a current source (from Aladdin's question) as a battery of unknown emf and variable internal resistance?

John Rennie

Current source

A current source is an electronic circuit that delivers or absorbs an electric current which is independent of the voltage across it. A current source is the dual of a voltage source. The term current sink is sometimes used for sources fed from a negative voltage supply. Figure 1 shows the schematic symbol for an ideal current source driving a resistive load. There are two types. An independent current source (or sink) delivers a constant current. A dependent current source delivers a current which is proportional to some other voltage or current in the circuit. == Background == An ideal current...

Guru Vishnu

08:41

@JohnRennie Thank you sir.

JR0005 : HCV-32-Ex-84

Question:
A capacitor of capacitance $C$ is given a charge $Q$. At $t=0$, it is connected to an ideal battery of emf $\epsilon$ through a resistance $R$. Find the charge on the capacitor at time $t$

Answer:
I found it to be
$$C\epsilon (1-e^{-t/RC})+Qe^{-t/RC}$$

and it is correct. I want to discuss some interesting inferences I obtained from the above expression, sir.

John Rennie

OK

Guru Vishnu

The first term represents the charge on a capacitor when it's charging and the second term represents the charge when it's discharging. In this question, we're adding both charging and discharging terms when there's already some charge on a capacitor to be charged completely.

From some question and answers on Physics and Electrical Engineering StackExchange, I learnt that a capacitor cannot be charged and discharged at the same time. But here it seems, both charging and discharging can take place in much the same way forward and backward reactions proceed in a chemical dynamic equilibrium.

John Rennie

It isn't really a combination of discharging and charging.

Guru Vishnu

Can a glass of water be filled when you are drinking out of it and pouring more water into the glass simultaneously? — Harry Svensson May 23 '18 at 23:11

John Rennie

It's a result of choosing the time origin.

Guru Vishnu

08:55

@JohnRennie Yes sir. We're shifting the time scale towards right. That's how I derived the above formula.

But it seems, it's safe to assume given two independent capacitors, one charged with $Q$ amount of charge and the other one completely discharged to discharge and charge simultaneously. And adding both or in other words superimposing the both cases, it seems the net result comes out to be the same.

I don't know whether I made the point clear. But I'm sure this is something similar to dynamic chemical equilibrium where both forward and reverse reactions take place.

John Rennie

I'm not sure these are comparable.

Guru Vishnu

desmos.com/calculator/rlrw079jak

John Rennie

In a reaction we reach an equilibrium state where the equilibrium is dynamic i.e. the rates of the forward and back reactions are the same.

And this really is a case where the reaction goes both ways.

In this problem the charge only ever flows in one direction.

Guru Vishnu

Ok sir. Now I understand why this case is different.

Did you see the graph? It's strikingly similar to the one I see in Chemistry.

John Rennie

Yes, but the similarity is superficial.

Guru Vishnu

09:04

Ok sir. Is there any additional inferences you could gather from this sir?

John Rennie

Nothing seems obvious to me ...

Guru Vishnu

Ok sir. Thank you.

End of #JR0005 : HCV-32-Ex-84

@JohnRennie: Are you free now sir?

John Rennie

09:22

@GuruVishnu yes

Guru Vishnu

I asked the following question on the main site on Jan 1 and received an answer. But still I have some doubts in that:

0

Q: Few conceptual doubts based on the effective capacitance of two metallic spheres joined by a metallic wire

The following problem is from Concepts of Physics by Dr. H.C.Verma, from the chapter "Capacitors", page 166, question 12: Problem: Two conducting spheres of radii $R_1$ and $R_2$ are kept widely separated from each other. What are their individual capacitances? If the spheres are connected b...

You may just read things under "My Conceptual Doubts:"

John Rennie

This isn't a spherical capacitor

Didn't we already talk about this?

Guru Vishnu

@JohnRennie We talked about spherical capacitors bigger plate to be at infinity having zero potential. The linked question, sir.

I think, but not about this one.

John Rennie

There are two types of capacitor. The one made from two plates, either parallel or concentric spheres, is called mutual capacitance.

However a single object also has a capacitance and this is called self capacitance.

Guru Vishnu

@JohnRennie Yes sir! I remember. And we obtained the same expression at last.

John Rennie

09:27

This question is about the self capacitance of the spheres.

Guru Vishnu

@JohnRennie Yes and I've assumed the self capacitance of individual spheres to be mutual capacitance of a bi-spherical capacitor with the larger sphere having infinitely large radius.

John Rennie

I don't think that's a good way to understand the capacitance.

Guru Vishnu

@JohnRennie Ok sir. Shall we proceed with self capacitance itself then? If so how do we define series and parallel combination? Just connect two "self(s)" and we have a parallel combination.

John Rennie

Imagine you have an isolated uncharged sphere.

It's potential is zero because it's uncharged.

Guru Vishnu

Ok sir.

John Rennie

09:33

Now suppose you add a charge $q$ to the sphere. It now has a potential. You know it has a potential because it is now surrounded by an electric field. We can calculate the potential very easily because the electric field around a spherical object is the same as the field from a point charge at the centre of the object.

OK so far?

Guru Vishnu

@JohnRennie Yes sir.

John Rennie

If the sphere has a radius $R$ then the potential at the surface is the same as the potential a distance $R$ from a point charge $q$, so it's $V = kq/R$.

Guru Vishnu

Yes sir.

John Rennie

Rearrange this to get $q = V \times R/k$ and compare with the capacitor equation $Q = CV$ and it's obvious the capacitance is $C= R/k = 4\pi\epsilon_0 R$.

Guru Vishnu

@JohnRennie Yes sir. Understood how we got the formula for self capacitance for each sphere. But what happens when we connect then using a wire?

John Rennie

09:39

When you connect the spheres with a wire you force them to be at the same potential. That means the charge distributes itself between the spheres to keep them at the same potential.

Guru Vishnu

@JohnRennie Then, I don't see what does capacitance mean when we do this.

John Rennie

Suppose the radii of the spheres are $R_1$ and $R_2$, and the charges on them are $Q_1$ and $Q_2$.

Guru Vishnu

I understood your point but the question is itself some kind of confusing on this part.

@JohnRennie Ok sir.

John Rennie

So $V_1 = Q_1/C_1 = Q_1/(4\pi\epsilon_0 R_1)$

Guru Vishnu

And similarly for $V_2$

John Rennie

09:41

And likewise $V_2 = Q_2/C_2 = Q_2/(4\pi\epsilon_0 R_2)$

Guru Vishnu

Ok sir.

John Rennie

And because the spheres are connected by a wire $V_1 = V_2$ so $Q_1/(4\pi\epsilon_0 R_1) = Q_2/(4\pi\epsilon_0 R_2)$

OK so far?

Guru Vishnu

@JohnRennie Ok. I don't think this will give us capacitance. May be I could be wrong.

The example earlier today became evident as you moved forward. But this seems to be a bit different.

John Rennie

And the total charge is $Q = Q_1 + Q_2$ so now we can solve for $Q_1$ and $Q_2$.

And the combined capacitance is just given by $C_T = Q/V$, where $V$ is the shared voltage $V_1 = V_2$.

Guru Vishnu

Ok sir. I'll try that way.

But aren't you tempted to think of mutual capacitance when the author writes "Think in terms of series-parallel connections."?

@JohnRennie: Hi sir.

John Rennie

09:49

I don't know what is meant by that. The question states the spheres are far apart, whcih I take to mean they are far enough apart that their electric fields don't influence each other i.e. they still behave as isolated spheres.

Guru Vishnu

@JohnRennie I think when they are said to be far apart, the charges on one small sphere doesn't have any impact on another small sphere.

But I don't see how it eliminates the case of the big sphere.

John Rennie

Yes

I'm now puzzled. What small and big spheres?

Guru Vishnu

@JohnRennie :-)

Spherical capacitor has two spheres - one small and another one big. The small one is concentric with the big one. And both are metallic.

John Rennie

By big sphere do you mean the hypothetical sphere at infinity?

Guru Vishnu

@JohnRennie Exactly!

@JohnRennie: Even though I found the correct answer using that method, I find this question to be confusing on the lines of realisation. Shall we quit this discussion for now?

John Rennie

09:57

I have to confess that I don't see how imagining a big sphere at infinity does anything except cause confusion.

Guru Vishnu

@JohnRennie When you find time, could you read the "My approach" section of my question. I've explained it there and I think it would be difficult to explain the same on chat. Is that fine sir?

Guru Vishnu

10:21

@JohnRennie: Acquisition of signal. Hi sir.

John Rennie

@GuruVishnu hi

Guru Vishnu

Did you read the question sir?

Or could you reply once you're free.

John Rennie

@GuruVishnu no. I have other things to do at the moment.

Guru Vishnu

@JohnRennie Ok sir. No problem. But could you comment on "My approach" when you find time? Thank you.

Guru Vishnu

12:05

@JohnRennie: Ok sir?

If it's not possible also, kindly let me know. Then I'll start a bounty.

John Rennie

@GuruVishnu I had a read through your approach, but I think it's the wrong approach so I remain unsurprised that it is leaving you puzzled about some things.

Guru Vishnu

12:53

Ok sir. Thank you.

Guru Vishnu

14:28

@JohnRennie: Hi sir. We know that speed of a particle remains constant when it is circulating in a uniform magnetic field. Is it possible to increase or decrease the speed using a non-uniform magnetic field?

Kindly reply when you find time.

I guess the charged particle would take an elliptical path rather than a circular one when the field is non-uniform. However, I don't see how to determine this mathematically.

John Rennie

@GuruVishnu no, because the Lorentz force is always normal to the velocity.

Guru Vishnu

@JohnRennie Thank you sir. So what happens in a non-uniform field? For sure it must not be circular because it lacks symmetry and defies some logic.

John Rennie

It won't be a circular path, but the speed will remain constant.

Guru Vishnu

@JohnRennie Ok sir. So could it be a open loop? A somewhat bent curve?

I don't see how could I do it for non-uniform fields.

John Rennie

It would depend on the field. In general it will be a complicated differential equation that probably doesn't have an analytic solution.

Guru Vishnu

14:41

I also tried some simulators, but all of them give results only for a uniform magnetic field.

John Rennie

The acceleration is the Lorentz force divided by the mass, and the acceleration is the derivative of the velocity, which is the derivative of the position vector, so we'd get:

$$ \frac{d^2\mathbf x}{dt^2} = \frac{q}{m} \frac{d\mathbf x}{dt} \times \mathbf B(\mathbf x) $$

Guru Vishnu

@JohnRennie Understood till this sir.

I think we need to define what a non-uniform field is.

Or how $B(x)$ varies with position.

John Rennie

Yes, you need to specify how $\mathbf B$ depends on $\mathbf x$.

Guru Vishnu

@JohnRennie Then shall we analyse it a bit qualitatively, sir?

And generalise the same for all kinds of field.

John Rennie

It's not going to be very exciting. It's just another differential equation that in general will be a pain to solve.

Guru Vishnu

14:48

@JohnRennie I agree sir. So shall we translate to the following question:

Is it not possible to alter the speed of a charged particle in a non-uniform magnetic field? If so how do Magnetic induction particle accelerators do their job (or is it totally a different concept; I guess that "induction" has something to do here)?

I think it's not possible as the magnetic force (Lorentz force) is always perpendicular to the instantaneous velocity vector.

John Rennie

@GuruVishnu correct

Guru Vishnu

@JohnRennie Ok sir. Thank you. What about the particle accelerator?

There wasn't a good diagram here - en.wikipedia.org/wiki/…; So I'm not sure whether it's magnetic field accelerating the charge or not.

John Rennie

4

A: Accelerating electrons via microwaves

Here's my fairly clumsy attempt to draw the sort of RF cavities used in particle accelerators. The usual disclaimer applies, my knowledge of accelerator design is strictly at the popular science level so caveat emptor. I'm sure the particle physicists hereabouts will pick me up on any errors. ...

Guru Vishnu

@JohnRennie Thank you for linking that answer. I think you've answers in all branches of physics all over the SE network.

22 secs ago, by John Rennie

@GuruVishnu you could try. Or contact Amazon tech support.

This is even more difficult to understand than that differential equation!

John Rennie

Oops, wrong chat room! :-)

Guru Vishnu

14:58

@JohnRennie :-) That was really funny however.

I just said for fun sir. I understood that Differential equation. It wasn't that difficult.

John Rennie

It's usually easy to write down the differential equation. A surprising number of them are just Newton's second law. It's solving the equation that's hard!

Guru Vishnu

@JohnRennie I think something more complicated is defining a non-uniform magnetic field. We're free to choose anything and don't know which one to use.

John Rennie

Yes.

Guru Vishnu

@JohnRennie: I think it's time for your lunch. Right?

John Rennie

In about five minutes! :-)

It's just cooking now.

Guru Vishnu

15:03

@JohnRennie Cool! I'm going to have my dinner after this discussion.

@JohnRennie Fine. Bye sir. Let's see tomorrow :-)

John Rennie

Bye :-)

Transcript for

♦ Particle Accelerator